We can use Newton's Law of Cooling: The temperature T of an object with intial temperature `T_0` after t minutes in a room with ambient temperature `T_r` is given by `T=Ce^(-kt)+T_r` .

Here `T_r=0` . We are given points `(t,T)` as `(0,100);(20,50)`

`100=Ce^(0t)==>C=100`

`50=100e^(-20k)`

`e^(-20k)=.5`

`-20k=ln(.5)`

`k=(ln(.5))/-20~~.035`

So `T=100e^(-.035t)` gives the temperature at time t:

(a) If the temperature of the bar is `25^@` find the time t:

`25=100e^(-.035t)`

`e^(-.035t)=.25`

`t=(ln(.25))/(-.035)~~39.6`

So the time required is approximately 39.6 minutes

(b) Find the temperature after 10 minutes:

`T=100e^(-.035*10)~~70.5`

So the temperature at 10 minutes is approximately `70.5^@`